(no subject)
So according to the "Psychrometric chart" or "Science" the relative humidity at normal pressure cannot be 100% at 92F so lets take the reference temp to be 80F that give us a mass ratio of about 0.022 lbs of water per lb of air so we will also need air and water densities at 80F.
62.22lbsH2O/ft^3
1 KgAir/m^3 --> 2.2lbsAir/35.3ft^3 --> 0.0623lbsAir/ft^3
desired volume of water was (16in*100mi*100mi)
convert to feet --> (1.33ft* 528000^2ft^2) --> 370,782,720,000 ft^3
so that is 23,070,100,838,400lbs or 11,535,050,419 tons
11.5 billion tons of water
te he he this is fun
so therefore for that much water you'll need 1,048,640,947,200,000lbs of Air
which is 16,832,117,932,584,270 ft^3
or 114,350 mi^3